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# Vogan圖之距離

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## 國 立 交 通 大 學

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### Abstract

A Vogan diagram is a Dynkin diagram with an involution, andthe vertices fixed by the involution may be black. If a Vogandiagram can be represented by another Vogan diagrams, then they are equivalent. Any Vogan diagram with many black vertices is equivalent to a diagram with only one black vertex. Our purpose is to find the steps from a Vogan diagram with many black vertices to one black vertex.

This thesis is divided into two parts. In the first part, consisting of Sections 1-5, we give a brief introduction of some fundamental concepts in Vogan diagram and the distance of two Vogan diagrams in classical types with proof. In the last part, Sections 6-9, we prove the distance of two Vogan diagrams in exceptional types.

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### 誌 謝

兩年的研究生生活，稍縱即逝，很感謝當初蔡孟傑教授願意當我的指導教授，並 且悉心教導，讓我能順利完成論文，除此之外，老師也會帶我們去打打羽球，培養感 情之餘順便磨練球技。感謝博士班的舉卿學姊，當課業上遇到問題，不論時間早晚， 她總是很有耐心教我，她也會告訴我哪裡有好玩好吃的，是位個性開朗、很照顧人的 學姐。 我很幸運有一群貼心的同學和室友，陪我分享一切，談談心事、談談未來，偶爾 也會一起打羽球，抒解身心，讓我的研究所生活一點也不孤單。我更要感激采瑩，她 從不吝於將自己所知道的教給我，也因為有她的鼓勵和打氣，讓我們能一塊完成論 文，一塊迎接口試過關的喜悅。謝謝凱奇學長教導我很多事情，除了論文，也給予我 精神上很大的支持。 求學路程一路走來，遇到再大的困難，都有家人在背後支撐著我，給我鼓勵，真 的很感謝媽媽、爸爸、哥哥、跟妹妹。我是個很戀家的人，即使家住南部仍常每個月 都會回家一次，媽媽為了不讓我坐車辛苦，偶爾會北上來看我，還帶很多好吃給我， 彼此聊著生活上的大小事，因為母親的相伴，讓我覺得讀書是件很幸福的事。 最後我將本論文的成果獻給我的父母、師長以及所有關心我的朋友，希望一起分 享完成論文的這份喜悅。

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### Introduction

A Dynkin diagram is a certain type of graph. It satisfies the property that two vertices may be connected by 0, 1, 2 or 3 edges, and an orientation (arrow) is assigned to each double or triple-edge. There are altogether seven classes of Dynkin diagrams, labelled as types A, B, C, D, E, F, G ([2, Chapter 11]). A Dynkin diagram in each type is specified by a subscript which indicates the number of vertices in that diagram, for example A5 is the diagram of type A which has 5 vertices. Then all the Dynkin

diagrams are given by An, Bn, Cn, Dn for n ≥ 1, as well as E6, E7, E8, F4, G2. The

infinite lists of diagrams of types A, B, C, D are called classical, and the finite lists of types E, F, G are called exceptional.

A Vogan diagram is a Dynkin diagram together with extra information. Namely, there is a diagram involution θ, such that the fixed points of θ are colored white or black [3]. Many Dynkin diagrams, for example Bn and Cn, have trivial symmetry. In

these cases the involution is the identity.

This thesis studies the following algorithm on the Vogan diagrams. Suppose that

v is a Vogan diagram, and p is a black vertex of v. Let Fp be the algorithm which

reverses the colors of all the θ-fixed vertices which are adjacent to p (but not p itself), except when the vertex is joint to p with a double-edge with arrow pointing towards

p. In this way, Fp(v) is another Vogan diagram. We give some examples as follows.

Example. The vertices are labelled 1, 2, . . . starting from the left.

v = f v v f ⇒ F2(v) = v v f f ⇒ F3(v) = f f v v u = f v f v=⇒v ⇒ F4(u) = f v v v=⇒f ⇒ F5(u) = f v f v=⇒v

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algorithms

v = v0 → v1 → ... → vk = w (1.1)

such that each step vi → vi+1 is given by some Fp. If v and w are equivalent and k in

(1.1) is as small as possible, we say that k = d(v, w) is the distance between v and w. Clearly a diagram without black vertex is not equivalent to any other one. Therefore, once and for all, we may consider only Vogan diagrams with black vertices, and denote them by V (X), where X is a Dynkin diagram with trivial diagram involution. Let

V1(X) ⊂ V (X) denote the diagrams with exactly one black vertex. It is known that

every diagram in V (X) is equivalent to a diagram in V1(X) [1][3]. It allows us to

define the distance between v ∈ V (X) and V1(X) by

d(v, V1(X)) = min

w∈V1(X)

d(v, w). (1.2) For fixed X, we intend to seek an upper bound for {d(v, V1(X)); v ∈ V (X)}, namely

d(V (X)) = max

v∈V (X)d(v, V1(X)). (1.3)

We present the main results of the thesis as follows. The classical Vogan diagrams are denoted by V (An), V (Bn), V (Cn), V (Dn), V (Dn, θ), where θ denotes the nontrivial

involution. We need not consider V (An, θ) because it contains only one diagram.

Theorem 1

(a) d(An) is bounded above by (n − 1)n2.

(b) d(Bn) is bounded above by (n − 1)n2.

(c) d(Cn) is bounded above by (n − 1)(n − 1)2.

(d) d(Dn) is bounded above by (n − 1)(n − 2)2.

(e) d(Dn, θ) is bounded above by (n − 1)(n − 2)2.

For the proof of Theorem 1, we shall study type A in Section 2, type B in Section 3, type C in Section 4, and type D in Section 5.

For the exceptional diagrams, only E6contains nontrivial involution θ. We provide

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Theorem 2

(a) d(E6) is bounded above by 130;

(b) d(E6, θ) is bounded above by 1;

(c) d(E7) is bounded above by 228;

(d) d(E8) is bounded above by 180;

(e) d(F4) is bounded above by 4;

(f ) d(G2) is bounded above by 1.

For the proof of Theorem 2, we shall study E6 in Section 6, E7 in Section 7, E8

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### Type A

Recall that a Dynkin diagram of type An is given by

h h p p p h h

1 2 n − 1 n

(2.1) A Vogan diagram of An is labelled as (2.1) with at least one black vertex, denoted

by V (An). According to the theorem by Borel and de Siebenthal[1, Theorem 6.96],

every Vogan diagram is equivalent to a Vogan diagram with 0 or 1 painted vertex. For type An, all Vogan diagrams with some black vertices can be equivalent to a painted

one. Let Vk(An) ⊂ V (An) denote the Vogan diagram of An with k black vertices.

From (1.2), it also extends to the distance between Vk(An) and V1(An) by

d(Vk(An), V1(An)) = min{d(v, w); v ∈ Vk(An), w ∈ V1(An)} (2.2)

or just write dk(V1(An)).

For convenience, we write (i1, . . . , ik) ∈ V (An) to denote the diagram with vertices

i1, . . . , ik painted. For example (3, 5) ∈ V (A6) means f f v f v f

In this section, we will discuss the distance between Vk(An) and V1(An); in other

words, we provide an upper bound for it.

d(V (An)) = max v∈V (An)

d(v, V1(An)). (2.3)

Example. For V (A4), there are 7 cases to discuss.

(1, 2) → (1)

(1, 3) → (1, 2, 3) → (2)

(1, 4) → (1, 2, 4) → (2, 3, 4) → (3) (2, 3) → (1, 2) → (1)

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(1, 2, 4) → (2, 3, 4) → (3)

(1, 2, 3, 4) → (2, 4) → (2, 3, 4) → (3)

By (2.3), d(V (A4)) = max{1, 2, 3, 2, 1, 2, 3} = 3.

Proposition 2.1 Let u = (i, j) ∈ V (An)

(a) If i − 1 ≤ n − j, then d(u, v) ≤ i(j − i) where v = (j − i) ∈ V (An).

(b) If i−1 > n−j, then d(u, w) ≤ (n−j+1)(j −i) where w = (n+i−j +1) ∈ V (An).

P roof : u has two black vertices in V (An), and it can be equivalent to a diagram in

V1(An); in other words, u ∼ s ∈ V1(A).

There are two ways to make u reduce to a Vogan diagram of An with only one

painted vertex:

Case 1 Move i and j towards the left:

e e p p p u p p p u p p p e 1 i j n e e p p p u p p p u p p p e 1 i − 1 j − 1 n e p p p u p p p u p p p p p p e 1 i − 2 j − 2 n p p p p p p u e p p p u e p p p e 1 j − i + 1 n p p p p p p e p p p u e p p p p p e j − i 1 n u = (i, j) −→ (i − 1, i, i + 1, j) −→ (i − 1, i + 1, i + 2, j) −→ · · · −→ (i − 1, j − 2, j − 1, j) −→ (i − 1, j − 1) needs at most j − i steps.

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Continuously,

(i − 1, j − 1) −→ (i − 2, i − 2) needs at most j − i steps. (i − 2, j − 2) −→ (i − 3, i − 3) needs at most j − i steps.

...

(2, j − i + 2) −→ (1, j − i + 1) needs at most j − i steps. Finally,

(1, j−i+1) −→ (1, 2, j−i+1) −→ (2, 3, j−i+1) −→ ··· −→ (j−i−1, j−i, j−i+1) −→ (j − i) needs at most j − i steps.

Thus, d(u, (j − i)) ≤ (j − i)[(i − 1) + 1] = i(j − i). Let v = (j − i) and d(u, v) = d1.

Case 2 Move i and j towards the right:

e e p p p u p p p u p p p e 1 i j n e p p p u p p p u p p p p p p e 1 i + 1 j + 1 n e p p p p p p u p p p u p p e 1 i + 2 j + 2 n p p p p p p e e p p p u e p p p u 1 n + i − j n p p p p p p e p p p p p p e u p p p e n + i − j + 1 1 n

u = (i, j) −→ (i, j − 1, j, j + 1) −→ (i, j − 2, j − 1, j + 1) −→ · · · → (i, i + 1, i + 2, j + 1) −→ (i + 1, j + 1) needs at most j − i steps.

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(i + 1, j + 1) −→ · · · −→ (i + 2, j + 2) needs at most j − i steps. (i + 2, j + 2) −→ · · · −→ (i + 3, j + 3) needs at most j − i steps.

...

(n + i − j − 1, n − 1) −→ (n + i − j, n) needs at most j − i steps. Finally,

(n + i − j, n) −→ (n + i − j, n − 1, n) −→ (n + i − j, n − 2, n − 1) −→ · · · −→ (n + i − j, n + i − j + 1, n + i − j + 2) −→ (n + i − j + 1) needs at most j − i steps. Thus,d(u, (n + i − j + 1)) ≤ (j − i)[(n + i − j + i) + 1] = (n − j + 1)(j − i). Let

w = (n + i − j + 1) and d(u, w) = d2.

Comparison with Case 1 and Case 2, the proof can be completed as the following: (a) i − 1 ≤ n − j ⇒ i < n − j + 1⇒ i(j − i) ≤ (n − j + 1)(j − i) ⇒ d1 ≤ d2 with

j − i > 0. By (1.2), it needs at most i(j − i) steps s.t. u → · · · → v = (j − i) ∈ V1(An).

(b) i − 1 > n − j ⇒ i > n − j + 1⇒ i(j − i) > (n − j + 1)(j − i) ⇒ d1 ≤ d2 with

j − i > 0. By (1.2), it needs at most (n − j + 1)(j − i) steps such that u → · · · → v = (j − i) ∈ V1(An).

Therefore, we can conclude that:

Proposition 2.2 Given u = (i1, i2, ..., ik) ∈ Vk(An), then dk(V1(An)) ≤ (k − 1)n2.

P roof : By Proposition 2.1, we have known that any two painted vertices which can

be reduced to be only one painted vertex in Vk(An) needs at most d1 or d2 steps.

Case 1 If i1 − 1 ≤ n − ik, and just move the leftmost two black vertices (i1, i2) in

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e p p p u p p p u u e p p p u e e p p p e

1 i1 i2 i3 ik n

o

e p p p p p p u p p p u e p p p u e e p p p e

1 i2− i1 i3 ik n

Let Dm = (im− im−1+ im−2− ... ± i1)(im+1− im+ im−1− im−2 + ... ∓ i1) where

0 < m < k. By Proposition 2.1(a), u ∼ (i2 − i1, i3, ..., ik) ∈ Vk−1 needs at most D1

steps.

Continuing the same way,

(i2− i1, i3, ..., ik) ∼ (i3 − i2 + i1, i4, ..., ik) ∈ Vk−2 needs at most D2 steps.

...

(ik−1− ik−2+ ... ± i1, ik) ∼ (ik− ik−1+ ... ∓ i1) ∈ V1 needs at most Dk−1 steps.

Then D1 ≤ i2· i2 = i22 ≤ n2, D2 ≤ i3· i3 = i32 ≤ n2, ... , Dk−1 ≤ ik· ik= ik2 ≤ n2,

and thus d(u, r) = D1 + D2 + ... + Dk−1 ≤ n2 + n2 + ... + n2 = (k − 1)n2, where

r = (ik− ik−1+ ... ∓ i1) ∈ V1.

Case 2 If i1− 1 > n − ik, and just move the rightmost two black vertices (ik−1, ik)

in the right direction:

e p p p e e u p p p e u u p p p u p p p e

1 i1 ik−2ik−1 ik n

o

e p p p e e u p p p e u p p p u p p p p p p e

1 i1 ik−2 n + ik−1− ik+ 1 n

By Proposition 2.1(b), u ∼ (i1, ..., ik−2, n + ik−1− ik+ 1) ∈ Vk−1 needs at most R1

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Continuously,

(i1, ..., ik−2, n + ik−1− ik+ 1) ∼ (i1, ..., ik−3, ik− ik−1+ ik−2) ∈ Vk−2 needs at most R2

steps where R2 = (ik− ik−1)(n + ik−1− ik+ 1 − ik−2).

...

(i1, ik− ik−1+ ik−2− ... ± i2) ∼ (n − ik+ ik−1− ... ∓ i2+ i1+ 1) ∈ V1 needs at most

Rk−1 steps where Rk−1 = (n − ik+ ik−1− ... ∓ i2+ i1+ 1)(ik− ik−1+ ik−2− ... ± i1).

Hence R1 ≤ n · n = n2, R2 ≤ n · n = n2, ... , Rk−1 ≤ n · n = n2.

Then d(u, t) = R1 + R2 + ... + Rk−1 ≤ n2 + n2 + ... + n2 = (k − 1)n2, where t =

(n − ik+ ik−1− ... ∓ i2+ i1+ 1) ∈ V1.

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### Type B

A Dynkin diagram of type Bn is given by

h h p p p h=⇒h

1 2 n − 1 n (3.1)

A Vogan diagram of Bn is labelled as (3.1) with at least one black vertex, denoted

by V (Bn). Let Vk(Bn) ⊂ V (Bn) denote the Vogan diagram of Bn with k black

vertices. From (1.2), it also extends to the distance between Vk(Bn) and V1(Bn) by

d(Vk(Bn), V1(Bn)) = min{d(v, w); v ∈ Vk(Bn), w ∈ V1(Bn)} (3.2)

or just write dk(V1(Bn)). Note that if n is painted, then Fn cannot reverse the color

of n − 1, or we can write Fn(w) = w, here w ∈ V (Bn)

In this section, we discuss the distance between Vk(Bn) and V1(Bn). We will

provide an upper bound for it.

Proposition 3.1 Let u = (i, i + k) ∈ V (Bn), then u ∼ v needs at most ki steps

where v = (k).

P roof : Suppose that we want to obtain a Vogan diagram of Bnwith one black vertex

from u immediately. Then we should move the painted vertices in the left direction, or the vertex n will be painted and it does not make sense that we mentioned before.

e p p p e u e p p p e e e p p p u p p p ee 1 i i + k n Fi e p p p u u u p p p e e e p p p u p p p ee 1 i − 1 i i + 1 i + k n Fi+1 e p p p u e u u p p p e e p p p u p p p ee 1 i − 1 i + 1 i + 2 i + k n F

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... ↓ Fi+k−2 e p p p u e e e p p p p p p e u u u p p p ee 1 i − 1 i + k − 2 → i + k − 1 i + k n Fi+k−1 e p p p u e e e p p p p p p e e u e p p p ee 1 i − 1 i + k − 1 n

The graph says that u = (i, i+k) −→ (i−1, i, i+1, i+k) −→ (i−1, i+1, i+2, i+k) −→

· · · −→ (i − 1, i + k − 2, i + k − 1, i + k) −→ (i − 1, i + k − 1) and it needs at most

[(i + k − 1) − i] + 1 = k steps. Continuously, we also can get that

(i − 1, i + k − 1) ∼ (i − 2, i + k − 2) needs at most k steps. (i − 2, i + k − 2) ∼ (i − 3, i + k − 3) needs at most k steps.

...

(2, 2 + k) ∼ (1, 1 + k) needs at most k steps.

Finally, (1, 1 + k) −→ (1, 2, 1 + k) −→ (2, 3, 1 + k) −→ · · · −→ (k − 1, k, k + 1) −→ (k) and it needs at least k steps.

Thus, there are at most k + [(i − 1) − 1]k + k = ki steps to make (i, i + k) ∼ (k); in other words, d(u, v) ≤ ki where v = (k).

Therefore, we can conclude that:

Proposition 3.2 Given u = (i1, i2, ..., ik) ∈ Vk(Bn), then dk(V1(Bn)) ≤ (k − 1)n2.

P roof : By Proposition 3.1, there are at most i(j − i) steps to make (i, j) reduced to

(j − i).

Let u = (i1, i2, ..., ik) ∈ Vk. Similar to Case 1 of Proposition 2.2, we move vertices

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and i2. Repeat this process of moving pairs of leftmost black vertices towards the

left. Finally, we prove that dk(V1(Bn)) ≤ (k − 1)n2. Throughout these steps, vertex

n remains unpainted.

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### Type C

The Vogan diagram of Cn is similar to Bnexcept for the direction of arrow on the

double-edge, it is indicated by (4.1) and denoted by V (Cn).

h h p p p h⇐=h

1 2 n − 1 n (4.1)

By Borel-Siebenthal Theorem, every Vogan diagram is equivalent to one with a black vertex. If a Vogan diagram belongs to V (Cn) and its vertex n is painted, then

this diagram is equivalent to (n). For example, (2, 4) ∈ V4(Cn) and then (2, 4) →

(2, 3, 4) → (3, 4) → (4).

Before talking about the distance between Vk(Cn) and V1(Cn), we consider the

distance between V2(Cn) and V1(Cn) first.

Proposition 4.1 Let u = (i, j) ∈ V (Cn).

(a) If j 6= n (i.e. n is unpainted), then

(i) d(u, v) ≤ i(j − i) if i − 1 ≤ (n − 1) − j, where v = (j − i).

(ii) d(u, w) ≤ (n − j)(j − i) if i − 1 ≥ (n − 1) − j, where w = (n − j + i). (b) If j = n (i.e. n is painted), then d(u, s) ≤ (n−i)(n−i+1)2 where s = (n).

P roof :

(a) If n is unpainted and we cannot use any method to make it become painted, then we can ignore the existence of n or just regard this Vogan diagram as V (An−1).

Recall the result of Proposition 2.1, then the proof of (i) and (ii) can be completed right away.

(b) If j = n, then u = (i, n).

Since n is black, there are no ways to let it change color. Now, we want to move the first black vertex closer to n(in other words, move it in the right direction), or they are too far to get the major distance.

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e p p p e u e e p p p e e eu 1 i n Fn e p p p e u e e p p p e e uu 1 i n − 1 n Fn−1 e p p p e u e e p p p e u uu 1 i n − 2 n − 1 n Fn−2 e p p p e u e e p p p u u eu 1 i n − 3 n − 2 n Fn−3 ... Fi+2 e p p p e u u u p p p e e eu 1 i i + 1 i + 2 n ↓ Fi+1 e p p p e e u e p p p e e eu 1 i + 1 n

The graph says that u = (i, n) −→ (i, n − 1, n) −→ (i, n − 2, n − 1, n) −→ (i, n − 3,

n−2, n) −→ · · · −→ (i, i+1, i+2, n) −→ (i+1, n) and it needs at most n−(i+1)+1

= n − i steps.

By the same way, we can get that

(i + 1, n) ∼ (i + 2, n) needs at most n − i − 1 steps. (i + 2, n) ∼ (i + 3, n) needs at most n − i − 2 steps.

...

(n − 2, n) ∼ (n − 1, n) needs at most 2 steps. (n − 1, n) −→ (n) needs 1 steps.

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= n−i X k=1 k = (n − i)(n − i − 1) 2 . Therefore, we can conclude that:

Proposition 4.2 Given u = (i1, i2, ..., ik) ∈ Vk(Cn). (a) If ik 6= n, then dk(V1(Cn)) ≤ (k − 1)(n − 1)2. (b) If ik = n, then dk(V1(Cn)) ≤ (k−1)n 2 2 . P roof :

(a) If ik6= n, then u ∈ V (An−1). Now, there are two cases to discuss:

i1− 1 ≤ (n − 1) − ik

i1− 1 > (n − 1) − ik

By Proposition 2.2, the result of (a) can be proved.

(b) First, we want to reduce the number of painted vertices step by step. With the same idea as Proposition 4.1(2), we move the first k − 1 black vertices closer to

n. Therefore, we just move ik−1 rightwards and make ik−1 and n reduce to one black

vertex. e p p p u e p p p u e u p p p e u p p p eu 1 i1 i2 i3 ik−1 n o e p p p u e p p p u e u p p p e e p p p eu 1 i1 i2 i3 n

Let Sp = 12(n − ik−p)(n − ik−p+ 1) where 0 < p < k. By Proposition 4.1(b),

u = (i1, i2, · · · , ik−1, n) ∼ (i1, i2, ik−2, · · · , n) ∈ Vk−1(Cn) needs at most S1 steps.

Continuously,

(i1, i2, · · · , ik−2, n) ∼ (i1, i2, ik−3, · · · , n) ∈ Vk−2(Cn) needs at most S2 steps.

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...

(i1, n) ∼ (n) ∈ V1(Cn) needs at most Sk−1 steps.

Then S1 12n · n = n 2 2 , S2 n 2 2 , S3 n 2 2 , ... , Sk−1 n 2 2 and thus dk(V1(Cn)) = S1+ S2+ · · · + Sk−1 n 2 2 (k − 1).

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### Type D

A Vogan diagram of Dn is a Dynkin diagram of Dn with a diagram involution θ,

denote it by V (Dn, θ)(see (5.1)). Besides, θ can be trivial and we also call it a Vogan

diagram of Dn, denote it by V (Dn) (see (5.2)). However the vertices n − 1 and n

fixed by θ can be painted.

h h p p p h¡ ¡ @ @ h h l θ 1 2 n − 2 n − 1 n (5.1) h h p p p h¡ ¡ @ @ h h 1 2 n − 2 n − 1 n (5.2) About (5.1) and (5.2), we give two examples to compare the difference between

V (Dn, θ) and V (Dn) if we use the same process.

Example : (1) e e u e u¡¡ @@ e e l θ −→F5 e e u u u¡¡ @@ e e l θ −→F4 e e e u e¡¡ @@ e e l θ

( i.e. (3, 5) ∈ V (D7, θ) and (3, 5) ∼ (4) needs 2 steps. )

(2) e e u e u¡¡ @@ e e −→F5 e e u u u¡¡ @@ u u −→F4 e e e u e¡¡ @@ u u −→F6 e e e u ¡ @@ u u −→F5 e e e e u¡¡ @@ e e

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( i.e. (3, 5) ∈ V (D7) and (3, 5) ∼ (5) needs 4 steps. )

Recall that Vk(Dn) means there are k black vertices in Vk(Dn). From (1.2), it also

extends to the distance between Vk(Dn) and V1(Dn) by

d(Vk(Dn), V1(Dn)) = min{d(v, w); v ∈ Vk(Dn), w ∈ V1(Dn)} (5.3)

or just write dk(V1(Dn)). In this section, we provide an upper bound for the distance

between Vk(Dn) and V1(Dn). Now, we discuss V (Dn, θ) as follows.

Proposition 5.1 Let u = (i, j) ∈ V (Dn, θ).

(a) If i − 1 ≤ (n − 2) − j, then d(u, v) ≤ i(j − i) where v = (j − i).

(b) If i − 1 > (n − 2) − j, then d(u, w) ≤ (n − j − 1)(j − i) where w = (n − j + i − 1).

P roof : Obviously, we can regard V (Dn, θ) as V (An−2) since n − 1 and n have no

colors. Then u = (i, j) ∈ V (An−2) and by Proposition 2.1, the results of (a) and (b)

here will be proved.

Therefore, we can conclude that:

Proposition 5.2 Given u = (i1, i2, ..., ik) ∈ Vk(Dn, θ), then dk(V1(Dn, θ)) ≤ (k − 1) ·

(n − 2)2.

P roof : Similarly, V (Dn, θ) is equivalent to V (An−2). Extending the result of

Propo-sition 2.2, we obtain that dk(V1(Dn, θ)) ≤ (k − 1)(n − 2)2 right away.

Consequently, we can get the result of Theorem 1 (e) right away. Continuously, we try to find the upper bound for d(Vk(Dn), V1(Dn)) and then discuss dk(V1(Dn)).

Proposition 5.3 Let u = (i, j) ∈ V (Dn).

(a) If j /∈ {n − 1, n}, then d(u, v) ≤ i(j − i) where v = (j − i).

(b) If i /∈ {n−1, n} and j ∈ {n−1, n}, then d(u, w) ≤ (n−i)(n−i−1)2 where w = (n−1) or (n).

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P roof :

(a) Suppose that j /∈ {n − 1, n}. Then i, j ∈ {1, 2, ..., n − 2}. Now, we want to

reduce u to be a Vogan diagram with one painted vertex. If we move these two black vertices toward right, then vertices n − 1 and n will become painted and the path to a Vogan diagram of V1(Dn) must be more complicated. Hence we move black vertices

toward left. e e p p p e u e p p p e e u p p p e¡¡ @@ e e i j ↓ Fi e e p p p u u u p p p e e u p p p e¡¡ @@ e e i − 1 i i + 1 j ↓ Fi+1 ... ↓ Fj−2 e e p p p u e e p p p u u u p p p e¡¡ @@ e e j − 2 i − 1 j ↓ Fj−1 e e p p p u e e p p p e u e p p p e¡¡ @@ e e i − 1 j − 1

It means that u = (i, j) ∼ (i − 1, j − 1) needs at most (j − i) steps. Continuously,

(i − 1, j − 1) ∼ (i − 2, j − 2) needs at most (j − i) steps. (i − 2, j − 2) ∼ (i − 3, j − 3) needs at most (j − i) steps.

...

(1, j − i + 1) ∼ (j − i) needs at most (j − i) steps.

So, there are at most i(j − i) steps to make (i, j) ∼ (j − i); in other words,

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(b) Suppose that i /∈ {n−1, n} and j /∈ {n−1, n}. Then u = (i, n−1) or u = (i, n),

0 < i < n−1. We only talk about u = (i, n−1) because (i, n) is equivalent to (i, n−1), and the conclusions are the same.

Now, we want to move these two black vertices closer and u will be reduced to a Vogan diagram of V1(Dn) as soon as possible.

e e p p p u e e p p p e e e¡¡ @@ u e i j = n − 1 ↓ Fn−1 e e p p p u e e p p p e e u¡¡ @@ u e i n − 2 j = n − 1 ↓ Fn−2 e e p p p u e e p p p e u u¡¡ @@ e u i n − 3 n ↓ Fn−3 e e p p p u e e p p p u u e¡¡ @@ e u i n − 4 n ↓ Fn−4 ... ↓ Fi+2 e e p p p u u u p p p e e e¡¡ @@ e u i i + 1 i + 2 n ↓ Fi+1 e e p p p e u e p p p e e e¡¡ @@ e u i + 1 n

We get (i, n − 1) ∼ (i + 1, n) needs at most (n − 1) − (i − 1) + 1 = n − i − 1 steps. By the same way, we also get

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(i + 1, n) ∼ (i + 2, n − 1) needs at most n − i − 2 steps. (i + 2, n − 1) ∼ (i + 3, n) needs at most n − i − 3 steps.

...

(n − 2, n − 1) ∼ (n) or (n − 2, n) ∼ (n − 1) needs at most 1 steps.

Thus, d(u, w) ≤ 1 + 2 + · · · + (n − i − 1) = (n−i)(n−i−1)2 where w = (n) or (n − 1). (c) We have known that vertices n − 1 and n are painted. The only way to let u be a diagram with one painted vertex is moving them in the left direction.

e e p p p e e p p p e e¡¡ @@ u u n − 1 n ↓Fn or Fn−1 e e p p p e e p p p e u¡¡ @@ u u n − 2 n − 1 n ↓Fn−2 e e p p p e e p p p u u¡¡ @@ e e n − 2 n − 1 n ↓Fn−3 ... ↓F2 u u p p p e e p p p e e¡¡ @@ e e 1 2 ↓F1 u e p p p e e p p p e e¡¡ @@ e e 1

From the above graph, it tells us that (n − 1, n) ∼ (1) needs at most (n − 1) − 1 + 1 steps, i.e. d(u, p) ≤ n − 1 where p = (1).

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Therefore, we can conclude that: Proposition 5.4 Given u = (i1, i2, ..., ik) ∈ Vk(Dn). (a) If n − 1, n /∈ u, then dk(V1(Dn)) ≤ (k − 1)(n − 2)2. (b) If n − 1 ∈ u or n ∈ u, then dk(V1(Dn)) ≤ (k−1)(n−1) 2 2 . (c) If n − 1, n ∈ u, then dk(V1(Dn)) ≤ (k − 3)(n − 2)2+ (n − 2). P roof :

(a) Suppose that 1 ≤ i1, i2, · · · , ik≤ n−2 (i.e. vertices n−1 and n are unpainted).

Looking back the method of Proposition 5.3(a), we just try to move the leftmost two black vertices toward left and then become one black vertex. Step by step, we can finally get only one black vertex in this Vogan diagram.

e p p p u e p p p u e u p p p u e p p p u e¡¡ @@ e e i1 i2 i3 ik−1 ik o e p p p e p p p u p p p u p p p u e p p p u e¡¡ @@ e e i2− i1 i3 ik−1 ik

Let Rm = (im− im−1+ im−2− ... ± i1)(im+1 − im+ im−1− im−2+ ... ∓ i1) where

0 < m < k. By Proposition 5.3(a), u ∼ (i2 − i1, i3, ..., ik) ∈ Vk−1(Dn) needs at most

R1 steps. Continuing the same way,

(i2− i1, i3, ..., ik) ∼ (i3 − i2 + i1, i4, ..., ik) ∈ Vk−2(Dn) needs at most R2 steps.

...

(ik−1− ik−2+ ... ± i1, ik) ∼ (ik− ik−1+ ... ∓ i1) ∈ V1(Dn) needs at most Rk−1 steps.

Then R1 ≤ i2· i2 = i22 ≤ (n − 2)2, R2 ≤ i3· i3 = i32 ≤ (n − 2)2,· · ·, Rk−1 ≤ ik· ik=

ik2 ≤ (n − 2)2.

Thus d(u, r) = R1+R2+...+Rk−1 ≤ (n−2)2+(n−2)2+...+(n−2)2 = (k−1)(n−2)2,

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(b) If n − 1 ∈ u or n ∈ u, then one of vertices n − 1 and n − 1 is painted (i.e.

ik = n − 1 or ik= n, but ik−1 6= n − 1).

Similar to the process of Proposition 5.3(b), we try to move the black vertex ik−1

towards the right. Finally, these two vertex will be reduced to a black vertex. Here we just talk about ik = n − 1 because its result is the same as ik = n.

e p p p u e p p p u e u p p p u u p p p e e¡¡ @@ u e i1 i2 i3 ik−2ik−1 ik = n − 1 o e p p p u e p p p u e u p p p u e p p p e e¡¡ @@ u e i1 i2 i3 ik−2 n − 1

Note that the rightmost black vertex could be n − 1 or n, but it does not affect the result. Let

Sp = 1

2(n − ik−p)(n − ik−p− 1), 1 ≤ p ≤ k − 1 . By Proposition 5.3(b),

u ∼ (i1, i2, ..., ik−2, n − 1) needs at most S1 steps. With the same method,

(i1, i2, ..., ik−2, n − 1) ∼ (i1, i2, ..., ik−3, n − 1) needs at most S2 steps.

(i1, i2, ..., ik−3, n − 1) ∼ (i1, i2, ..., ik−4, n − 1) needs at most S3 steps.

...

(i1, n − 1) ∼ (n − 1) needs at most Sk−1 steps.

We find that S1 12(n−2)(n−1), S2 12(n−2)(n−1), · · · , Sk−1 12(n−2)(n−1).

Thus d(u, w) = S1+ S2+ · · · + Sk−1 12(k − 1)(n − 2)(n − 1), where w = (n − 1) or

(n) and this proof is completed.

(c) Let ik−1 = n − 1 and ik = n, i.e. u = (i1, i2, ..., ik−2, n − 1, n). Regardless of

vertices n − 1 and n, we hope to reduce the k − 2 black vertices in 1, · · · , n − 1 to one black vertex. Then adding in vertices n − 1 and n, we finally make these three black vertices become only one black vertex.

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Referring to the method of (a), we move the leftmost two painted vertices towards the left. e p p p u e p p p u e u p p p u u p p p e e¡¡ @@ u u i1 i2 i3 ik−2 n − 1 n o e p p p e e u p p p e u p p p u u p p p e e¡¡ @@ u u i2− i1 i3 ik−2 n − 1 n o ... o e p p p e e e p p p e e p p p u e p p p e e¡¡ @@ u u n − 1 n

Let Ms = (is− is−1 + · · · ± i1)(is+1− is+ is−1− · · · ∓ i1) where 1 ≤ s ≤ k − 3.

By Proposition 5.3(a), u ∼ (i2− i1, i3, . . . , n − 1, n) needs at most M1 steps. Keep on

the same way:

(i2− i1, i3, . . . , n − 1, n) ∼ (i3− i2+ i1, i4, . . . , n − 1, n) needs at most M2 steps.

(i3− i2 + i1, i4, . . . , n − 1, n) ∼ (i4 − i3 + i2 − i1, i5, . . . , n − 1, n) needs at most M3

steps.

...

(ik−3− ik−4+ ... ± i1, ik−2, n − 1, n) ∼ (ik−2− ik−3+ ... ∓ i1, n − 1, n) needs at most

Mk−3 steps.

Let a = ik−2 − ik−3 + ... ∓ i1 and the final step is to find vertex i such that

(a, n − 1, n) ∼ (i):

(a, n − 1, n) −→ (a, n − 2, n − 1, n) −→ (a, n − 3, n − 2) −→ (a, n − 4, n − 3) −→

· · · −→ (a, a + 1, a + 2) −→ (a + 1) needs at most (n − a − 1) steps and i = a + 1.

We have known that 1 ≤ a ≤ n − 2, then 1 ≤ n − a − 1 ≤ n − 2.

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≤ i1· i2+ i2 · i3 + · · · + ik−3· ik−2+ (n − a − 1)

≤ (n − 2)2+ (n − 2)2+ · · · + (n − 2)

= (k − 3)(n − 2)2+ (n − 2).

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### Type E

6

After observing An, Bn, Cnand Dn, we talk about other different type: E6, E7, E8, F4

and G2. e e e e e e E6 e e e e e e e E7 e e e e e e e e E8 e e=⇒e e F4 e=⇒e G2 (6.1)

This section will deal with the upper bound for the distance between Vk(E6) and

V1(E6). Others are discussed later. Besides (6.1), there is another type of E6, unfixed

by θ, denoted by V (E6, θ).

e e e e e e

θ

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e e e e e e 1 2 3 4 5 0 e e e e e e 1 2 3 4 5 0 θ

We give an example about V (E6) and V (E6, θ) on how to reduce some black

vertices to only one black vertex.

Example : V (E6): u e u e u e −→F3 u u u u u u −→F2 e u e u u u −→F4 e u u u e u −→F3 e e u e e e i.e. (1,3,5)−→(0,1,2,3,4,5)−→(0,2,4,5)−→(0,2,3,4)−→(3) V (E6, θ): e e u e e u θ −→F0 e e e e e u θ i.e. (0,3)−→(0)

Indeed, there are two vertices in V (E6, θ) can be drawn color, vertex 0 and vertex

3, just like the vertices n and n − 1 in V (Dn, θ).

Recall that

dk(V1(E6)) = d(Vk(E6), d1(E6)) = min{d(u, v); u ∈ Vk(E6), v ∈ V1(E6)}

Proposition 6.1 dk(V1(E6)) ≤ 5(5k − 4) where 1 ≤ k ≤ 6.

P roof : Separate E6 into A5 and a vertex 0. Regardless of vertex 0, we reduce other

black vertices to be one black vertex in V (A5). There are 4 cases to discuss.

Case 1 Consider vertex 0 is white and all black vertices are in vertex 1 to 3 or vertex 3 to 5. Then vertex 0 will not become painted. By Proposition 2.2, we

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can get the distance between Vk(E6) to V1(E6) directly. dk(V1(E6)) = dk(V1(A5)) ≤

52(k − 1) = 25(k − 1).

Case 2 Consider vertex 0 is black and all black vertices are in vertex 1 to 3 or vertex 3 to 5. Then vertex 0 will not become unpainted. We arrange the black vertices in V (A5) to be singly white and finally deal with vertex 0. We have to discuss

the following cases.

(i) u ∼ (0, 1) and (0, 1) −→ (0, 1, 2) −→ (0, 2, 3) −→ (3, 4) −→ (4, 5) −→ (5). By Proposition 2.2, we have known that dk(V1(A5)) ≤ 52(k − 1) = 25(k − 1). Then

d(u, (5)) ≤ 25(k − 1) + 5 = 25k − 20.

(ii) u ∼ (0, 2) and (0, 2) −→ (0, 2, 3) −→ (3, 4) −→ (4, 5) −→ (5). Then d(u, (5)) ≤ 25(k − 1) + 4 = 25k − 21.

(iii) u ∼ (0, 3) and (0, 3) −→ (0). Then d(u, (5)) ≤ 25(k − 1) + 1 = 25k − 24.

We don’t handle with (0, 4) and (0, 5) because they are equivalent to (0, 2) and (0, 1). Thus, dk(V1(E6)) ≤ max{25k − 20, 25k − 21, 25k − 24} = 25k − 20 where

1 ≤ k ≤ 6.

Case 3 Consider vertex 0 is white. Some black vertices are in vertex 1 to 3 and others are in vertex 3 to 5. Then the process of becoming singly painted will affect the color of vertex 0. We let vertex 0 be black and use the same method as Case 2. Similarly, there are 3 cases to discuss and thus dk(V1(E6)) ≤ 25k − 20 where

1 ≤ k ≤ 5.

Case 4 Consider vertex 0 is black. Some black vertices are in vertex 1 to 3 and others are in vertex 3 to 5. The color of vertex 0 will be reversed in the process of reducing to singly painted vertex. We let vertex 0 be white and regard V (E7) as

V (A6). Thus, dk(V1(E6)) = dk(V1(A5)) ≤ 25(k − 1) = 25k − 25 where 1 ≤ k ≤ 6.

Arrange above 4 cases, we get that dk(V1(E6)) ≤ 25k − 20 = 5(5k − 4) where

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Proposition 6.2 dk(V1(E6, θ)) ≤ 1 where k = 2.

P roof : Since vertices 1, 2, 4 and 5 are unfixed by θ, we do not have to paint any

color on them. In other words, only vertices 0 and 3 can have color. Hence, the only one case is like the above example, (0,3)−→(0), needs only one step. Therefore, we get the proof of Theorem 2(b) immediately.

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### Type E

7

By (6.1), we also label 0,1,...,7 to the Vogan diagram of E7.

e e e e e e e

1 2 3 4 5 6 0

The process of making some black vertices to be singly painted is similar to V (E6).

Now, we discuss how many steps that Vk(E7) goes to V1(E7) at most.

Proposition 7.1 Given u = (i1, i2, ..., ik) ∈ Vk(E7), dk(V1(E7)) ≤ 12(3k − 2) where

1 ≤ k ≤ 7.

P roof : Separate E7 into A6 and a vertex 0. We try to discuss the distance between

k painted vertices and only one painted vertex by the color of vertex 0.

Case 1 Suppose that vertex 0 is white and all black parts are in vertices 1 to 3 or vertices 3 to 6. Then the process of reducing to only one black vertex cannot affect the color of vertex 0. Thus we just regard V (E7) as V (A6) and by Proposition 2.2,

dk(V1(E7)) = dk(V1(A6)) ≤ 62(k − 1) = 36k − 36.

Case 2 Suppose that vertex 0 is black and other black parts are in vertices 1 to 3 or vertices 3 to 6. Then the process of reducing to only one black vertex cannot affect the color of vertex 0. We make these painted vertices in V (A6) become one

painted and then act on vertex 0. There are 6 cases to be discussed.

(i) u ∼ (0, 1) and (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6). By Proposition 2.2, we have known that dk(V1(A6)) ≤ 62(k − 1) = 36(k − 1).

Then d(u, (6)) ≤ 36(k − 1) + 6 = 36k − 30.

(ii) u ∼ (0, 2) and (0, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6). Then d(u, (6)) ≤ 36(k − 1) + 5 = 36k − 31.

(iii) u ∼ (0, 3) and (0, 3) → (0).

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(iv) u ∼ (0, 4) and (0, 4) → (0, 3, 4) → (2, 3) → (1, 2) → (1). Then d(u, (1)) ≤ 36(k − 1) + 4 = 36k − 32. (v) u ∼ (0, 5) and (0, 5) → (0, 3, 5) → (2, 3, 4, 5) → (2, 4) → (1, 2, 3, 4) → (0, 1, 3) → (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6). Then d(u, (6)) ≤ 36(k − 1) + 12 = 36k − 24. (vi) u ∼ (0, 6) and (0, 6) → (0, 5, 6) → (0, 4, 5) → (0, 3, 4) → (2, 3) → (1, 2) → (1). Then d(u, (1)) ≤ 36(k − 1) + 6 = 36k − 30. Thus, dk(V1(E7)) ≤ max{36k−30, 36k−31, 36k−35, 36k−32, 36k−24, 36k−30} = 36k − 24 where 1 ≤ k ≤ 7.

Case 3 Suppose that vertex 0 is white and some black vertices are in vertex 1 to 3 and the other black in vertex 3 to 6. The color of vertex 0 will be reversed in the process of reducing to singly painted vertex. We let vertex 0 be black and use the same method as Case 2. Similarly, there are 6 cases to discuss and thus

dk(V1(E7)) ≤ 36k − 24 where 1 ≤ k ≤ 6.

Case 4 Suppose that vertex 0 is black and some black vertices are in vertex 1 to 3 and others are in vertex 3 to 6. The color of vertex 0 will be reversed in the process of reducing to singly painted vertex. We let vertex 0 be white and look V (E7) as

V (A6). Thus, dk(V1(E7)) = dk(V1(A6)) ≤ 36(k − 1).

Combining with Case 1 to Case 4, we can get that dk(V1(E7)) ≤ 36k − 24 =

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### Type E

8

From the diagram of (6.1), we have known that what the Vogan diagram of E8

looks like. Similarly, we give 8 numbers on E8.

e e e e e e e e

1 2 3 4 5 6 7 0

This section also talk about what is the maximal distance between Vk(E8) and

V1(E8).

Proposition 8.1 Given u = (i1, i2, ..., ik) ∈ Vk(E8), then dk(V1(E8)) ≤ 5(5k − 4)

where 1 ≤ k ≤ 8.

P roof : Use the same way as Proposition 6.1 and Proposition 7.1, we also have to

talk about 4 cases. As above, we just calculate Case 2 and Case 3 since they need more steps.

I. Suppose that vertex 0 is black and other black parts are in vertices 1 to 3 or vertices 3 to 7. Then the process of reducing to only one black vertex cannot affect the color of vertex 0. We make these painted vertices in V (A7) become one painted

and then act on vertex 0. There are 7 cases to be discussed.

(i) u ∼ (0, 1) and (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) → (7).

By Proposition 2.2, we have known that dk(V1(A7)) ≤ 72(k − 1) = 49(k − 1).

Then d(u, (7)) ≤ 49(k − 1) + 7 = 49k − 42. (ii) u ∼ (0, 2) and (0, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) → (7). Then d(u, (7)) ≤ 49(k − 1) + 6 = 49k − 43. (iii) u ∼ (0, 3) and (0, 3) → (0). Then d(u, (0)) ≤ 49(k − 1) + 1 = 49k − 48. (iv) u ∼ (0, 4) and (0, 4) → (0, 3, 4) → (2, 3) → (1, 2) → (1).

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Then d(u, (1)) ≤ 49(k − 1) + 4 = 49k − 45. (v) u ∼ (0, 5) and (0, 5) → (0, 3, 5) → (2, 3, 4, 5) → (2, 4) → (1, 2, 3, 4) → (0, 1, 3) → (0, 1) → (0, 1, 2) → (0, 2, 3) → (3, 4) → (4, 5) → (5, 6) → (6, 7) → (7). Then d(u, (7)) ≤ 49(k − 1) + 13 = 49k − 36. (vi) u ∼ (0, 6) and (0, 6) → (0, 3, 6) → (2, 3, 4, 6) → (2, 4, 5, 6) → (2, 5) → (1, 2, 3, 5) → (0, 1, 3, 4, 5) → (0, 1, 4) → (0, 1, 3, 4) → (1, 2, 3) → (2). Then d(u, (1)) ≤ 49(k − 1) + 10 = 49k − 39. (vii) u ∼ (0, 7) and (0, 7) → (0, 6, 7) → (0, 5, 6) → (0, 4, 5) → (0, 3, 4) → (2, 3) → (1, 2) → (1). Then d(u, (1)) ≤ 49(k − 1) + 7 = 49k − 42. Thus, dk(V1(E8)) ≤ max{49k − 42, 49k − 43, 49k − 48, 49k − 45, 49k − 36, 49k − 39, 49k − 42} = 49k − 36 where 1 ≤ k ≤ 8.

II. Suppose that vertex 0 is white and some black vertices are in vertex 1 to 3 and the other black in vertex 3 to 7. The color of vertex 0 will be reversed in the process of reducing to singly painted vertex. We let vertex 0 be black and use the same method as I. Similarly, there are 7 cases to discuss and thus dk(V1(E8)) ≤ 49k − 36 where

1 ≤ k ≤ 7.

Combining with I and II, we can get that dk(V1(E8)) ≤ 49k −36 where 1 ≤ k ≤ 8.

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4

### and G

2

Finally, we look the remaining Vogan diagram, F4 and G2 and observe that the

upper bound for d(Vk(F4), V1(F4)) (k = 2, 3, 4) and d(Vk(G2, V1(G2)).

Proposition 9.1 dk(V1(F4)) ≤ 4 where k = 2, 3, 4.

P roof : Since there are only 4 vertices in V (F4), we just use diagram to find the

distance between Vk(F4) and V1(F4), k = 2, 3, 4.

u u=⇒e e−→ u e=⇒e e u e=⇒u e−→ u u=⇒u e−→ e u=⇒e e u e=⇒e u−→ u u=⇒e u−→ u u=⇒u u−→ u u=⇒u e−→ e u=⇒e e e u=⇒u e−→ u u=⇒e e−→ u e=⇒e e e u=⇒e u−→ u u=⇒u u−→ u u=⇒u e−→ e u=⇒e e e e=⇒u u−→ e e=⇒e u u u=⇒u e−→ e u=⇒e e u u=⇒e u−→ u u=⇒u u−→ u u=⇒u e−→ e u=⇒e e u e=⇒u u−→ u u=⇒u u−→ u u=⇒u e−→ e u=⇒e e e u=⇒u u−→ e u=⇒u e−→ u u=⇒e e−→ u e=⇒e e u u=⇒u u−→ u u=⇒u e−→ e u=⇒e e

From the above process, we can find that it needs at most 4 steps to make a Vogan diagram with some black vertices reduce to only one black vertex.

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Proposition 9.2 dk(V1(G2)) ≤ 1 where k = 2.

P roof : V (G2) is graphed as e=⇒eand there are at most 2 vertices to be painted.

So, we just talk about the only one case, (1, 2).

u=⇒u−→ u=⇒e

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### References

[1] A. Borel and J. de Siebenthal, Les sous-groupes fermes de rang maximum des groupes de Lie clos, Comment. Math. Helv. 23 (1949), 200-221.

[2] J. E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer-Verlag, New York 1972.

(a) In a compact metric space X, every sequence contains a subsequence converging to a point in X, i.e... Then X has the leat-upper-bound property if and only if every Cauchy

For 5 to be the precise limit of f(x) as x approaches 3, we must not only be able to bring the difference between f(x) and 5 below each of these three numbers; we must be able

remember from Equation 1 that the partial derivative with respect to x is just the ordinary derivative of the function g of a single variable that we get by keeping y fixed.. Thus

The Vertical Line Test A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once.. The reason for the truth of

– Maintain a heavy path: Instead of recalculate all ances tors' value, only the corresponding overlapping subpa th will be recalculated. It cost O(1) time for each verte x, and

Proof: For every positive integer n, there are finitely many neighbor- hood of radius 1/n whose union covers K (since K is compact). Collect all of them, say {V α }, and it forms

 Students are expected to explain the effects of change in demand and/or change in supply on equilibrium price and quantity, with the aid of diagram(s). Consumer and producer

If x or F is a vector, then the condition number is defined in a similar way using norms and it measures the maximum relative change, which is attained for some, but not all